The angle between tangents to the parabola `y^2=4ax` at the points where it intersects with teine `x-y-a = 0` is `(a> 0)`

Options

A. `pi//3`
B. `pi//4`
C. `pi//6`
D. `pi//2`

Explanation

To solve the problem of finding the angle between the tangents to the parabola \( y^2 = 4ax \) at the points where it intersects with the line \( x - y - a = 0 \), we can follow these steps:

### Step 1: Identify the parabola and the line
The given parabola is \( y^2 = 4ax \) and the line is \( x - y - a = 0 \).

### Step 2: Find the points of intersection
To find the points of intersection, we can substitute \( y \) from the line equation into the parabola equation.

From the line equation, we can express \( y \) as:
\[ y = x - a \]

Now substitute \( y \) into the parabola equation:
\[ (x - a)^2 = 4ax \]

Expanding this gives:
\[ x^2 - 2ax + a^2 = 4ax \]

Rearranging the equation:
\[ x^2 - 6ax + a^2 = 0 \]

### Step 3: Solve the quadratic equation
We can solve the quadratic equation \( x^2 - 6ax + a^2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -6a, c = a^2 \):
\[ x = \frac{6a \pm \sqrt{(-6a)^2 - 4 \cdot 1 \cdot a^2}}{2 \cdot 1} \]
\[ x = \frac{6a \pm \sqrt{36a^2 - 4a^2}}{2} \]
\[ x = \frac{6a \pm \sqrt{32a^2}}{2} \]
\[ x = \frac{6a \pm 4a\sqrt{2}}{2} \]
\[ x = 3a \pm 2a\sqrt{2} \]

### Step 4: Find corresponding \( y \) values
Now we substitute back to find the corresponding \( y \) values:
1. For \( x = 3a + 2a\sqrt{2} \):
\[ y = (3a + 2a\sqrt{2}) - a = 2a + 2a\sqrt{2} = 2a(1 + \sqrt{2}) \]
2. For \( x = 3a - 2a\sqrt{2} \):
\[ y = (3a - 2a\sqrt{2}) - a = 2a - 2a\sqrt{2} = 2a(1 - \sqrt{2}) \]

### Step 5: Find the slopes of the tangents
The slope of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_0, y_0) \) is given by:
\[ m = \frac{2a}{y_0} \]

Calculating the slopes for both points:
1. For \( (3a + 2a\sqrt{2}, 2a(1 + \sqrt{2})) \):
\[ m_1 = \frac{2a}{2a(1 + \sqrt{2})} = \frac{1}{1 + \sqrt{2}} \]
2. For \( (3a - 2a\sqrt{2}, 2a(1 - \sqrt{2})) \):
\[ m_2 = \frac{2a}{2a(1 - \sqrt{2})} = \frac{1}{1 - \sqrt{2}} \]

### Step 6: Find the angle between the tangents
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Calculating \( m_1 m_2 \):
\[ m_1 m_2 = \frac{1}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{1}{1 - 2} = -1 \]

Thus:
\[ \tan \theta = \left| \frac{\frac{1}{1 + \sqrt{2}} - \frac{1}{1 - \sqrt{2}}}{1 - 1} \right| \]
Since \( m_1 m_2 = -1 \), the tangents are perpendicular, hence:
\[ \theta = 90^\circ \]

### Conclusion
The angle between the tangents to the parabola at the points of intersection with the line is \( 90^\circ \).